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University of South Carolina - Steven Derocher's Math241 topics covered, focused on 13.1-13.4 and any additional formulas given

1

(a=alpha) horizontal component of a projectile

x0+v0cos(a)t

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2

(a=alpha) vertical component of a projectile

y0+v0sin(a)t-1/2(g)t^2

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3

a (alpha)=

the angle formed by the projectile

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4

the gravitational constant, g, equals

9.8

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5

to solve for the horizontal distance of a projectile

set the y component equal to 0, solve for t, then plug this value into the x component

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6

(a=alpha) ideal projectile motion from point (x0,y0), r=

(x+|v0|(cosa)t)i+(y+|v0|(sina)t-(1/2)gt^2)j

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7

v0 is the

initial velocity of the projectile

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8

length of a smooth curve (r=x(t)i+y(t)j+z(t)k), L=

∫a→b sqrt((dx/dt)^2+(dy/dt)^2+dz/dt)^2)dt

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9

arc length formula (simplified)

∫a→b|v(t)|dt

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10

(T=tau) arc length parameter with base point P(t0), s(t)=

∫t0→t sqrt((x’T)^2+(y’T)^2+(z’T))dT

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11

(T=tau) simplified arc length parameter, s(t)=

∫t0→t|v(T)|dT

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12

unit tangent vector, T(t)=

v(t)/|v(t)|, the direction the particle is travelling in

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13

speed on a smooth curve

ds/dt=|v(t)|

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14

(k=kappa) curvature formula 1 (good for understanding) k=

|dT/ds|

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15

(k=kappa, T=unit tangent vector) second curvature formula, k=

(1/|r’(t)|)(dT/dt)

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16

principal unit normal vector, N=

(dT/dt)/(|dT/dt|), the direction T(t) is changing in

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17

(k=kappa) third curvature formula (the one we typically use), k=

(|T(t)|X|N(t)|)/(|T(t)|)^3

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18

radius of an osculating circle

1/k

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19

bynormal vector, B=

TXN

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20

TNB frame

<T(t), N(t), B(t)>, for osculating circles

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21

k of a circle is…, k of a line is…

constant, 0

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22

r(t) is orthogonal to r’(t) if

r(t) is differentiable and |r(t)| is constant

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23

(d/dt)(u(f(t)))=f’(t)(u’(f(t))

define u as <g(t), h(t)>, plug in f(t) for t, apply (d/dt)<g(f(t)), h(f(t))> using chain rule, factor and simplify

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24

(projectile motion) to find distance travelled by projectile,…

set j component equal to 0, solve for t (throw out any t’s that don’t make sense), plug t into r(t), solve

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25

to find max height of projectile,…

solve for t where r’(t)=0, check that it is a max using 2nd derivative test (left side should be positive, right side should be negative), plug this t into r(t)

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